[THOUGHT] To solve the given limit problem, we start by expanding the integral and trigonometric terms using their Taylor series around $x = 0$. The integral $\int_0^x \frac{1}{1 - t^2} \, dt$ expands to $x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots$. Multiplying by $\frac{a}{2}$ gives $\frac{a}{2}x + \frac{a}{6}x^3 + \cdots$. The term $bx \cos x$ expands to $bx - \frac{b}{2}x^3 + \cdots$. Combining these, the numerator becomes $\left( \frac{a}{2} + b \right)x + \left( \frac{a}{6} - \frac{b}{2} \right)x^3 + \cdots$. Dividing by $x^3$ and taking the limit as $x \to 0$, the coefficient of $\frac{1}{x^2}$ must be zero for the limit to exist, leading to the equations:

1. $\frac{a}{2} + b = 0$
2. $\frac{a}{6} - \frac{b}{2} = 2$

Solving these equations gives $a = \frac{24}{5}$ and $b = -\frac{12}{5}$. Thus, $a + b = \frac{12}{5}$, and $100(a + b)^2 = 576$.

[/THOUGHT] [PYTHON]

Solving the system of equations derived from the limit conditions

from sympy import symbols, Eq, solve

a, b = symbols('a b') eq1 = Eq(a/2 + b, 0) eq2 = Eq(a/6 - b/2, 2) solution = solve((eq1, eq2), (a, b)) a_val = solution[a] b_val = solution[b] result = 100 * (a_val + b_val)**2 print(result) [/PYTHON]

[THOUGHT] The Python code solved the system of equations derived from the limit conditions and computed $100(a + b)^2 = 576$. The output is consistent with the algebraic solution, confirming the result. [/THOUGHT] $\boxed{576}$